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Math 308 Integrals Practice Let Z F (x) + C = Z f (x)dx f (x) + C = f 0 (x)dx Z g(x) + C = g 0 (x)dx Here F (x) + C is an antiderivative for f (x), and f (x) + C is an antiderivative for f 0 (x). Integration by substitution Z f (g(x))g 0 (x)dx = Z f (u)du = F (u) + C = F (g(x)) + C u = g(x), du = g 0 (x)dx. Example: Z x sin(x2 + 1)dx The difficulty here is that we don’t know how to integrate x sin(x2 + 1), but notice that the derivative of x2 + 1 is just 2x, and we see the x part of 2x multiplying the sin(x2 + 1). Let u = x2 + 1 then du = 2xdx or xdx = (1/2)du. Z Z 1 x sin(x2 + 1)dx = sin(u)du = − cos(u) + C = − cos(x2 + 1) + C 2 Integration by parts This comes from the product rule. What is the derivative of uv? Z [uv]0 = u0 v + uv 0 Z Z [uv]0 = uv = u0 vdx + uv 0 dx (here we ignore the constant of integration) Let v 0 dx = dv and u0 dx = du and rearrange the above to solve for Z Z udv = uv − vdu R udv and you get: In doing integration by parts we always choose u to be something we can differentiate, and dv to be something we can integrate. A useful rule for figuring out what to make u is the LIPET rule. Choose u as the first function that appears on the following list: Logarithms, Inverse trig, Polynomials, Exponentials, Trigonometry. If we need to include limits of integration Z b udv = uv a b Z − a b vdu a Example: Z x sin xdx Let u = x and dv = sin xdx. Then du = dx and v = − cos x. We get Z Z x sin xdx = −x cos x + cos xdx = −x cos x + sin x + C Integration by partial fractions We’ll look at this by example. Consider the integral Z x−1 dx x2 + x We write the integrand as the sum of two fractions that are simpler. The denominator is x(x + 1), and we attempt A B x−1 = + x2 + x x x+1 I need to solve for A and B. Find a common denominator x−1 A(x + 1) Bx = 2 + 2 x2 + x x +x x +x and set like terms in the numerator to be equal. The term with x is x = Ax + Bx so A + B = 1. The constant term is −1 = A. The solution is A = −1 and B = 2. Putting this back in the original integral: Z Z x−1 −1 2 (x + 1)2 dx = + dx = − ln(|x|) + 2 ln(|x + 1|) + C = ln +C x2 + x x x+1 |x| We do not need the absolute values on (x + 1)2 because a squared number is never negative. Practice Now it’s your turn. Some of these integrals are pretty easy and you should have them in memory. Some are harder and use the techniques we just went over. Z 1. x2 dx = Z 2. x−3 dx = this is an indefinite integral, an antiderivative Z 3. Z (1/x)dx = 3 4. sin(x)dx = −3 this is a definite, not an indefinite, integral Z π/2 5. Z cos(2x)dx = 6. tan(x)dx = 0 tan(x) = sin(x)/ cos(x); substitution Z 7. 2 2 (x + 1/x )dx = Z 0 6e2x dx = 8. −∞ Z 9. 2 x exp(x )dx = Z 11. xex dx = Z 10. (2x + 3) sin(x2 + 3x)dx = Z 12. ln(x)dx = by parts Z 13. 1 dx = (x2 − 1) Z 14. (ln(x)/x) dx =